All Solutions
Page 619: Practice Problems
R = $R_{1} + R_{2} + R_{3}$
R = 20 + 20 + 20
$boxed{R = 60 Omega}$
In this question, we have three resistors. Each resistor is of 20 $Omega$.
So the formula will be as follow:
R = $R_{1} + R_{2} + R_{3}$
$mathbf{I = V / R}$
I = 120 / 60
$$
boxed{I = 2.0 text{A}}
$$
$mathbf{I = V /R}$
Where:
I = current ,
V = voltage and
R = resistance.
(b) $boxed{I = 2.0 text{A}}$
Equivalent resistance $R_e$ of three resistors connected in series is calculated as:
$$
begin{align*}
R_e &= R_1 + R_2 + R_3 \
tag{in our case all resistors are the same $R_1 = R_2 = R_3 = R$} \
R_e &= R + R + R \
R_e &= 3 R \
tag{plug in the values} \
R_e &= 3 cdot 20 ~Omega
end{align*}
$$
$$
boxed{ R_e = 60 ~Omega }
$$
To find the current $I$ in the circuit, we first need to understand that since these resistors are connected in series, this current will be the same current $I$ flowing through the generator. Current $I$ through the battery can be calculated from Ohm’s law, by using the equivalent resistance $R_e$ of all the resistors in the circuit:
$$
begin{align*}
I &= dfrac{epsilon}{R_e} \
I &= dfrac{ 120 ~mathrm{V} }{60 ~Omega }
end{align*}
$$
$$
boxed{ I = 2 ~mathrm{A} }
$$
R_e = 60 ~Omega
$$
$$
I = 2 ~mathrm{A}
$$
$R_2 = 15 ~Omega$ and $R_3 = 5 ~Omega$ connected in series across a battery with electromotive force of $epsilon = 90 ~mathrm{V}$. We must find equivalent resistance $R_e$ of the these three resistors connected in series and current $I$ in the circuit.
Equivalent resistance $R_e$ of three resistors connected in series is calculated as:
$$
begin{align*}
R_e &= R_1 + R_2 + R_3 \
tag{plug in the values} \
R_e &= 10 ~Omega + 15 ~Omega + 5 ~Omega
end{align*}
$$
$$
boxed{ R_e = 30 ~Omega }
$$
To find the current $I$ in the circuit, we first need to understand that since these resistors are connected in series, this current will be the same current $I$ flowing through the battery. Current $I$ through the battery can be calculated from Ohm’s law, by using the equivalent resistance $R_e$ of all the resistors in the circuit:
$$
begin{align*}
I &= dfrac{epsilon}{R_e} \
I &= dfrac{ 90 ~mathrm{V} }{30 ~Omega }
end{align*}
$$
$$
boxed{ I = 3 ~mathrm{A} }
$$
R_e = 30 ~Omega
$$
$$
I = 3 ~mathrm{A}
$$
R = $R_{1} + R_{2} + R_{3}$
R = 10 + 5 + 15
$$
boxed{R = 30 Omega}
$$
a) The resistors in this circuit are connected in series. When the resistors are connected in series, you can calculate the total resistance by adding their resistance together.
In this question, we have three resistors.
$R_{1}$ = 10 $Omega$
$R_{2}$ = 15 $Omega$
$R_{3}$ = 5 $Omega$
So the formula will be as follow:
R = $R_{1} + R_{2} + R_{3}$
I = V / R
I = 90 / 30
$$
boxed{I = 3.0 text{A}}
$$
(b) As the resistors are connected in series, the flow of the current will be the same throughout the circuit. The voltage in the circuit is 90 V and the total resistance is 30 $Omega$
So to calculate the current in the circuit, the formula is as follows:
$mathbf{I = V /R}$
Where:
I = current
V = voltage
R = resistance.
(b) $boxed{I = 3.0 text{A}}$
$$R_e=R_1+R_2+R_3$$
We can see that equivalent resistance will **increase** by the **same amount **by which we increased resistance of one of the resistors.
$$I=frac{V}{R_e}$$
Which means that by increasing equivalent resistance, we will **decrease** current in the circuit.
$R_2$ and $R_3$ connected in series across a $V = 9 ~mathrm{V}$ battery.
$$
a)~~
$$
Equivalent resistance $R_e$ of three resistors connected in series is calculated as a sum of individual resistances of resistors:
$$
begin{align*}
R_e &= R_1 + R_2 + R_3 tag{1}
end{align*}
$$
We see that since equivalent resistance $R_e$ of resistors connected in series is equal to a sum of individual resistances of resistors, if we increase resistance of either of the resistors, equivalent resistance $R_e$ increases. We can also understand this as equivalent resistance $R_e$ being proportional to the resistance of each individual resistor connected in series, which means that increasing the resistance of either of the resistors leads to an increase in $R_e$.
$b)~~$ Current $I$ in the circuit could be found from Ohm’s law, but we first need to understand that since these resistors are connected in series, this current will be the same current $I$ flowing through the battery. Current $I$ through the battery can be calculated from Ohm’s law, by using the equivalent resistance $R_e$ of all the resistors in the circuit:
$$
begin{align*}
I &= dfrac{V}{R_e} tag{2}
end{align*}
$$
We see that current $I$ through the circuit is inversely proportional to the equivalent resistance $R_e$ of resistors connected in series. As said, increase in resistance of either of the resistors will lead to increase in equivalent resistance $R_e$, which in turn will lead to a decrease in current.
$c)~~$ Voltage $V$ across the battery used above is actually electromotive force $epsilon$ of the battery and electromotive force of the battery doesn’t depend on the change of resistors in the circuit nor the change of current in the circuit. Emf of the battery is same if it’s connected in the circuit and when it’s not connected to any circuit.
$b)~~$ Current decreases.
$c)~~$ Electromotive force of the battery doesn’t change.
$$R_e=10cdot R$$
$$I=frac{epsilon }{R_e}$$
$$R_e=frac{120}{0.06}$$
Finally:
$$boxed{R_e=2000,,rm{Omega }}$$
$$R_e=10cdot R$$
$$R=frac{R_e}{10}$$
$$R=frac{2000}{10}$$
$$boxed{R=200,,rm{Omega }}$$
$$R=200,,rm{Omega }$$
$epsilon = 120 ~mathrm{V}$, current $I$ flowing through the bulbs is
$I = 0.06 ~mathrm{A}$. Essentially, we have 10 resistors with same resistance $R$ connected in series. Equivalent resistance $R_e$ of resistors connected in series is equal to a sum of individual resistances, stated as:
$$
begin{align*}
R_e &= R + R + R + R + R + R + R + R + R + R \
R_e &= 10 R tag{1} \
end{align*}
$$
$a)~~$ Current $I$ in this circuit will be the same current $I$ flowing through the outlet. This current can be calculated from Ohm’s law, by using the equivalent resistance $R_e$ of all the resistors in the circuit:
$$
begin{align*}
I &= dfrac{epsilon}{R_e} \
tag{express $R_e$ from the equation} \
R_e &= dfrac{epsilon}{I} \
tag{plug in the values} \
R_e &= dfrac{ 120 ~mathrm{V} }{ 0.06 ~mathrm{A} }
end{align*}
$$
$$
boxed{ R_e = 2000 ~mathrm{Omega } }
$$
$b)~~$ To find resistance of each individual bulb in the holiday lights, let’s go back to equation $(1)$ from which we can calculate resistance $R$ of each individual bulb:
$$
begin{align*}
R_e &= 10 R \
tag{express $R$ from equation above} \
R &= dfrac{R_e}{10 } \
R &= dfrac{ 2000 ~mathrm{Omega } }{10 }
end{align*}
$$
$$
boxed{ R = 200 ~mathrm{Omega } }
$$
a)~~ R_e = 2000 ~mathrm{Omega }
$$
$$
b)~~ R = 200 ~mathrm{Omega }
$$
$$V=IR$$
Calculating for each resistor we get:
$$V_1=IR_1$$
$$V_2=IR_2$$
$$V_3=IR_3$$
$$V_1=3cdot 10$$
$$V_2=3cdot 15$$
$$V_3=3cdot 5$$
$$boxed{V_1=30,,rm{V}}$$
$$boxed{V_2=45,,rm{V}}$$
$$boxed{V_3=15,,rm{V}}$$
$$epsilon=V_1+V_2+V_3$$
Inserting values we get:
$$90=30+45+15$$
$$90=90$$
Which means that:
$$boxed{epsilon=V_1+V_2+V_3 }$$
$$V_2=45,,rm{V}$$
$$V_3=15,,rm{V}$$
$$epsilon=V_1+V_2+V_3 $$
$I = 3 ~mathrm{A}$. All three resistors with resistances $R_1 = 10 ~Omega$,
$R_2 = 15 ~Omega$ and $R_3 = 5 ~Omega$ are connected in series to a battery with electromotive force of $epsilon = 90 ~mathrm{V}$.
To prove that voltage drops $V_1$, $V_2$ and $V_3$ across resistors $R_1$, $R_2$ and $R_3$ add up to emf of the battery $epsilon$, we’ll apply Ohm’s law. Note that the same current $I$ will flow through all of the resistors because they’re connected in series. Ohm’s law states that current $I$ flowing through a resistor with resistance $R$ and voltage $V$ across it is equal to:
$$
begin{align*}
I &= dfrac{V}{R} \
tag{express $V$ from the equation above} \
V &= I R tag{1}
end{align*}
$$
This means that we can apply equation $(1)$ to calculate voltage drop $V$ on each of the resistors as:
$$
V_1 = I R_1 = 3 ~mathrm{A} cdot 10 ~Omega = 30 ~mathrm{V}
$$
$$
V_2 = I R_2 = 3 ~mathrm{A} cdot 15 ~Omega = 45 ~mathrm{V}
$$
$$
V_3 = I R_3 = 3 ~mathrm{A} cdot 5 ~Omega = 15 ~mathrm{V}
$$
As we can see, if we add up the voltage drops $V_1$, $V_2$ and $V_3$ we have:
$$
begin{align*}
V_1 + V_2 + V_3 &= 30 ~mathrm{V} + 45 ~mathrm{V} + 15 ~mathrm{V} \
V_1 + V_2 + V_3 &= 90 ~mathrm{V}
end{align*}
$$
Notice $epsilon = 90 ~mathrm{V}$. We have proven that
$$
boxed{ V_1 + V_2 + V_3 = epsilon }
$$
$$
V = I R
$$
$R_A = 255 ~Omega$ and
$R_B =292 ~Omega$ connected in series. We’re also told that voltage across resistor $R_A$ is
$V_A = 17 ~mathrm{A}$.
$$
I = dfrac{V }{R}
$$
We can apply Ohm’s law to resistor $R_A$. Voltage across this resistor is $V_A$ and current flowing through it is $I$. We thus write:
$$
begin{align*}
I &= dfrac{V_A}{R_A} \
I &= dfrac{ 17 ~mathrm{A} }{ 255 ~Omega }
end{align*}
$$
$$
boxed{ I = 0.06667 ~mathrm{A} }
$$
$$
I = dfrac{V}{ R_e}
$$
Since resistors $R_A$ and $R_B$ are connected in series, their equivalent resistance is
$$
R_e = R_A + R_B
$$
We can plug in the expression for $R_e$ into the equation above and have:
$$
begin{align*}
I &= dfrac{V}{R_A + R_B} \
tag{express $V$ from the equation above} \
V &= I (R_A + R_B) \
tag{plug in the values} \
V &= 0.06667 ~mathrm{A} cdot ( 255 ~Omega + 292 ~Omega )
end{align*}
$$
$$
boxed{ V = 36.4667 ~mathrm{V} }
$$
$$
P = I^2 R
$$
Since the same current $I$ flows through both resistor $R_A$ and $R_B$, we find that power dissipation $P_A$ on resistor $R_A$ is:
$$
begin{align*}
P_A &= I^2 R_A \
tag{plug in the values} \
P_A &= ( 0.06667 ~mathrm{A} )^2 cdot 255 ~Omega
end{align*}
$$
$$
boxed{ P_A = 1.1333 ~mathrm{W} }
$$
Power dissipation $P_B$ on resistor $R_B$ is:
$$
begin{align*}
P_B &= I^2 R_B \
tag{plug in the values} \
P_B &= ( 0.06667 ~mathrm{A} )^2 cdot 292 ~Omega
end{align*}
$$
$$
boxed{P_B = 1.2978 ~mathrm{W} }
$$
Total power dissipation $P$ is power dissipated on the equivalent resistor $R_e$, which means:
$$
begin{align*}
P &= I^2 R_e \
tag { plug in $R_e = R_A + R_B $ } \
P &= I^2 (R_A + R_B) tag{1} \
tag{plug in the values} \
P &= (0.06667 ~mathrm{A} )^2 ( 255 ~Omega + 292 ~Omega )
end{align*}
$$
We have calculated that total power dissipation $P$ is equal to:
$$
boxed{ P = 2.4311 ~mathrm{W} }
$$
$$
begin{align*}
P_A + P_B &= 1.1333 ~mathrm{W} + 1.2978 ~mathrm{W} \
P_A + P_B &= 2.4311 ~mathrm{W} \
tag{notice that $ P = 2.4311 ~mathrm{W} $ } \
P_A + P_B &= P
end{align*}
$$
We notice that power dissipation $P_A$ and $P_B$ on individual resistors add up to total power dissipation. This is always the case with resistors connected in any type of connection. We can prove this by going back to equation $(1)$:
$$
begin{align*}
P &= I^2 (R_A + R_B) \
tag{distribute the terms in the brackets} \
P &= I^2 R_A + I^2 R_B \
tag{notice $P_A = I^2 R_A $ } \
tag{notice $P_B = I^2 R_B$ } \
P &= P_A + P_B
end{align*}
$$
Clearly, when two resistor are connected in series, total power dissipation $P$ is equal to a sum of individual power dissipations $P_A$ and $P_B$.
$$
dfrac{1}{R_e} = dfrac{1}{R_A} + dfrac{1}{R_B}
$$
For simplicity, power $P$ dissipated on resistor $R$ with voltage $V$ across it can be calculated as:
$$
P = dfrac{V^2}{R}
$$
We’ll use this equation for power dissipation $P$ in parallel connection because it includes voltage $V$ across the resistor, and since resistors $R_A$ and $R_B$ are connected in parallel, voltage across them is the same. Note that this is also the voltage $V$ across the battery.
Power $P_A$ dissipated on resistor $R_A$ is thus calculated as:
$$
P_A = dfrac{V^2}{R_A}
$$
Power $P_B$ dissipated on resistor $R_B$ is calculated as:
$$
P_ B = dfrac{V^2}{R_B}
$$
$$
begin{align*}
P &= dfrac{V^2}{R_e} \
tag{rewrite the equation as follows} \
P &= V^2 cdot dfrac{1}{R_e} \
tag{notice that $dfrac{1}{R_e} = dfrac{1}{R_A} + dfrac{1}{R_B} $} \
P &= V^2 cdot left( dfrac{1}{R_A} + dfrac{1}{R_B} right) \
tag{distribute the term in the brackets} \
P &= dfrac{V^2}{R_A} + dfrac{V^2}{R_B} \
tag{notice that $P_A = dfrac{V^2}{R_A} $} \
tag{notice that $P_ B = dfrac{V^2}{R_B}$ } \
P &= P_A + P_B
end{align*}
$$
Now that we have proven that total power $P$ is equal to a sum of power dissipated on individual resistors, no matter the type of connection between those two resistors, we conclude that the equation
$$
P = P_1 + P_2 + P_3 + …
$$
will follow for any type of connection between any resistors, where $P_1$, $P_2$ … are power dissipations on the resistors, connected in any way.
begin{align*}
a)~~ I &= 0.06667 ~mathrm{A} \
\
b)~~ V &= 36.4667 ~mathrm{V} \
\
c)~~ P_A &= 1.1333 ~mathrm{W} \
c)~~ P_B &= 1.2978 ~mathrm{W} \
c)~~ P &= 2.4311 ~mathrm{W}
end{align*}
$$
Hint: Total power dissipation $P$ is calculated as power dissipated on the equivalent resistor and it’s always equal to a sum of powers dissipated on individual resistors.
We also need to discuss why some sets might blow their fuses after many bulbs have failed.
$$I=frac{V}{R_e}$$
we can see that by lowering the total resistance, the current in the circuit will increase. When it reaches a certain treshold it will result in a failure of the complete set.
$$
R_A < R_B
$$
At first, when these two resistors are off and there is no current flowing through them, temperature on them is the same and let's call it $T_0$. When current $I$ starts flowing through the resistors, since these two resistors are connected in series, same current $I$ will flow through both of them. This means that power output $P_A$ on resistor $R_A$ will be equal to:
$$
P_A = I^2 R_A
$$
whereas power output $P_B$ on resistor $R_B$ will be equal to:
$$
P_B = I^2 R_B
$$
During the time interval $t$ , energy output $E_A$ on resistor $R_A$ will be equal to:
$$
E_A = P_A t = I^2 R_A t
$$
whereas energy output $E_B$ on resistor $R_B$ will be equal to:
$$
E_B = P_B t = I^2 R_B t
$$
$$
begin{align*}
Q = m c Delta T
end{align*}
$$
where $m$ is mass of resistor $R$, $c$ is specific heat capacity from which resistor is made from and $Delta T = T – T_0$ is change in the temperature of the resistor. Assuming that mass of both of the resistors is the same $m_A = m_B = m$ and that they’re made from the same material, which results in them having the same specific heat capacity $c_A = c_B = c$ we find that heat output $Q_A$ on resistor $R_A$ during the time $t$ is equal to:
$$
Q_A = m c Delta T_A = mc (T_A – T_0)
$$
where $T_A$ is temperature to which resistor $R_A$ heats up. We also find that heat output $Q_B$ of resistor $R_B$ during the time $t$ is equal to:
$$
Q_B = m c Delta T_B = mc (T_B – T_0)
$$
where $T_B$ is temperature to which resistor $R_B$ heats up.
$$
Q_A = E_A ~~~ rightarrow ~~~ I^2 R_A t = mc (T_A – T_0)
$$
$$
Q_B = E_B ~~~ rightarrow ~~~ I^2 R_B t = mc (T_B – T_0)
$$
We can now divide the two equations above and have:
$$
begin{align*}
dfrac{ I^2 R_A t}{ I^2 R_B t } &= dfrac{mc (T_A – T_0)}{mc (T_B – T_0) } \
tag{cancel out $I^2, m, c$} \
dfrac{R_A}{R_B} &= dfrac{T_A – T_0}{T_B – T_0 } \
tag{notice that $R_A < R_B ~~rightarrow~~ dfrac{R_A}{R_B} < 1 $} \
dfrac{R_A}{R_B} &= dfrac{T_A – T_0}{T_B – T_0 } < 1 \
dfrac{T_A – T_0}{T_B – T_0 } &< 1 \
tag{multiply by $(T_B – T_0$} \
T_A – T_0 &< T_B – T_0 \
tag{cancel out $ T_0 $} \
T_A &< T_B
end{align*}
$$
As we can see, if resistance $R_A < R_B$, temperature $T_A$ of resistor $R_A$ will be lower than the temperature $T_B$ of resistor $R_B$.
$$epsilon=V_1+V_2+V_3$$
$$V_3=epsilon – V_1 – V_2$$
$$V_3=12-1.21-3.33$$
Voltage on the third resistor is:
$$boxed{V_3=7.46,,rm{V}}$$
$$
V = V_1 + V_2 + V_3
$$
where $V$ is voltage across the battery (its emf) and $V_1$, $V_2$ and $V_3$ is voltage across resistors $R_1$, $R_2$ and $R_3$, respectively. We can now use equation above to calculate voltage $V_3$ across the third resistor:
$$
begin{align*}
V &= V_1 + V_2 + V_3 \
tag{express $V_3$ from the equation above } \
V_3 &= V – V_1 – V_2 \
tag{plug in the given values} \
V_3 &= 12 ~mathrm{V} – 1.21 ~mathrm{V} – 3.33 ~mathrm{V} \
end{align*}
$$
$$
boxed{ V_3 = 7. 46 ~mathrm{V} }
$$
V_3 = 7. 46 ~mathrm{V}
$$
a)~~
$$
In this problem we have two resistors with resistances $R_1 = 22 ~Omega$,
$R_2 = 33 ~Omega$ connected in series across a battery with voltage
$V = 120 ~mathrm{V}$. We must find equivalent resistance $R_e$ of the these two resistors connected in series and current $I$ in the circuit.
Equivalent resistance $R_e$ of two resistors connected in series is calculated as:
$$
begin{align*}
R_e &= R_1 + R_2 \
tag{plug in the values} \
R_e &= 22~Omega + 33 ~Omega
end{align*}
$$
$$
boxed{ R_e = 55 ~Omega }
$$
b)~~
$$
To find the current $I$ in the circuit, we first need to understand that since these resistors are connected in series, the same current $I$ will flow through each of the resistors and through the battery. Current $I$ through the battery can be calculated from Ohm’s law, by using the equivalent resistance $R_e$ of all the resistors in the circuit:
$$
begin{align*}
I &= dfrac{V}{R_e} \
I &= dfrac{ 120 ~mathrm{V} }{ 55 ~Omega }
end{align*}
$$
$$
boxed{ I = 2.1818 ~mathrm{A} }
$$
$$
I = dfrac{V}{R}
$$
where $V$ is voltage drop across this resistor. We see that we can express voltage drop $V$ from the equation above as:
$$
V = I R
$$
which means that voltage drop $V$ across resistor $R$ can be calculated as a product of resistance $R$ of the resistor and current $I$ flowing through it.
This means that voltage drop $V_1$ across resistor $R_1$ is equal to:
$$
begin{align*}
V_1 &= I R_1 \
tag{plug in the values} \
V_1 &= 2.1818 ~mathrm{A} cdot 22 ~Omega
end{align*}
$$
$$
boxed{ V_1 = 48~mathrm{V} }
$$
and that voltage drop $V_2$ across resistor $R_2$ is equal to:
$$
begin{align*}
V_2 &= I R_2 \
tag{plug in the values} \
V_2 &= 2.1818 ~mathrm{A} cdot 33 ~Omega
end{align*}
$$
$$
boxed{ V_2 = 72 ~mathrm{V} }
$$
d)~~
$$
Voltage drop across the two resistors together is equal to:
$$
begin{align*}
V_1 + V_2 &= 48~mathrm{V} + 72 ~mathrm{V} \
V_1 + V_2 &= 120 ~mathrm{V} \
tag{notice $V = 120 ~mathrm{V} $}
end{align*}
$$
$$
boxed{ V_1 + V_2 = V }
$$
As expected, voltage drop across each individual resistor connected in series add up to voltage $V$ across the battery.
begin{align*}
& a)~~ R_e = 55 ~Omega \
& b)~~ I = 3 ~mathrm{A} \
& c)~~ V_1 = 48~mathrm{V} \
& c)~~ V_2 = 72 ~mathrm{V} \
& d) ~~ V_1 + V_2 = V = 120 ~mathrm{V}
end{align*}
$$
a)~~
$$
In this problem we have three resistors with resistance $R_1 = 3.3 ~mathrm{k Omega }$,
$R_2 = 4.7 ~mathrm{k Omega }$ and $R_3 = 3.9 ~mathrm{k Omega }$ connected in series across a battery with voltage $V = 12 ~mathrm{V}$.
Equivalent resistance $R_e$ of three resistors connected in series is calculated as:
$$
begin{align*}
R_e &= R_1 + R_2 + R_3 \
tag{plug in the values} \
R_e &= 3.3 ~mathrm{k Omega } + 4.7 ~mathrm{k Omega } + 3.9 ~mathrm{k Omega }
end{align*}
$$
$$
boxed{ R_e = 11.9 ~mathrm{k Omega } }
$$
b)~~
$$
To find the current $I$ in the circuit, we first need to understand that since these resistors are connected in series, this current will be the same current $I$ flowing through the battery. Current $I$ through the battery can be calculated from Ohm’s law, by using the equivalent resistance $R_e$ of all the resistors in the circuit:
$$
begin{align*}
I &= dfrac{V}{R_e} \
tag{plug in the values} \
I &= dfrac{ 12 ~mathrm{V} }{11.9 ~mathrm{k Omega } } \
tag{$1 ~mathrm{k Omega } = 1000 ~mathrm{Omega} $} \
I &= dfrac{ 12 ~mathrm{V} }{11.9 cdot 1000 ~mathrm{ Omega } } \
I &= 1.0084 cdot 10^{-3} ~mathrm{A}
end{align*}
$$
$$
boxed{ I = 1.0084 ~mathrm{mA} }
$$
$$
I = dfrac{V}{R}
$$
where $V$ is voltage drop across this resistor. We see that we can express voltage drop $V$ from the equation above as:
$$
V = I R
$$
which means that voltage drop $V$ across resistor $R$ can be calculated as a product of resistance $R$ of the resistor and current $I$ flowing through it.
This means that voltage drop $V_1$ across resistor $R_1$ is equal to:
$$
begin{align*}
V_1 &= I R_1 \
tag{plug in the values} \
V_1 &= 1.0084 ~mathrm{mA} cdot 3.3 ~mathrm{k Omega } \
tag{$1 ~mathrm{k Omega } = 1000 ~mathrm{Omega} $} \
tag{$ 1 ~mathrm{mA} = 10^{-3} ~mathrm{A} $} \
V_1 &= 1.0084 cdot 10^{-3} ~mathrm{A} cdot 3.3 cdot 1000 ~mathrm{ Omega } \
V_1 &= 1.0084 ~mathrm{A} cdot 3.3 ~mathrm{ Omega }
end{align*}
$$
$$
boxed{ V_1 = 3.32772 ~mathrm{V} }
$$
$$
begin{align*}
V_2 &= I R_2 \
tag{plug in the values} \
V_2 &= 1.0084 ~mathrm{mA} cdot 4.7 ~mathrm{k Omega } \
tag{$1 ~mathrm{k Omega } = 1000 ~mathrm{Omega} $} \
tag{$ 1 ~mathrm{mA} = 10^{-3} ~mathrm{A} $} \
V_2 &= 1.0084 cdot 10^{-3} ~mathrm{A} cdot 4.7 cdot 1000 ~mathrm{ Omega } \
V_2 &= 1.0084 ~mathrm{A} cdot 4.7 ~mathrm{ Omega }
end{align*}
$$
$$
boxed{ V_2 = 4.73948 ~mathrm{V} }
$$
and voltage drop $V_3$ across resistor $R_3$ is equal to:
$$
begin{align*}
V_3 &= I R_3 \
tag{plug in the values} \
V_3 &= 1.0084 ~mathrm{mA} cdot 3.9 ~mathrm{k Omega } \
tag{$1 ~mathrm{k Omega } = 1000 ~mathrm{Omega} $} \
tag{$ 1 ~mathrm{mA} = 10^{-3} ~mathrm{A} $} \
V_3 &= 1.0084 cdot 10^{-3} ~mathrm{A} cdot 3.9 cdot 1000 ~mathrm{ Omega } \
V_3 &= 1.0084 ~mathrm{A} cdot 3.9 ~mathrm{Omega }
end{align*}
$$
$$
boxed{ V_3 = 3.93276 ~mathrm{V} }
$$
d)~~
$$
Voltage drop across the three resistors together is equal to:
$$
begin{align*}
V_1 + V_2 + V_3 &= 3.32772 ~mathrm{V} + 4.73948 ~mathrm{V} +3.93276 ~mathrm{V} \
V_1 + V_2+ V_3 &= 12 ~mathrm{V} \
tag{notice $V = 12 ~mathrm{V} $}
end{align*}
$$
$$
boxed{ V_1 + V_2 + V_3 = V }
$$
As expected, voltage drop across each individual resistor connected in series add up to voltage $V$ across the battery.
begin{align*}
&a)~~ R_e = 11.9 ~mathrm{k Omega } \
&b)~~ I = 1.0084 ~mathrm{mA} \
& c)~~ V_1 = 3.32772 ~mathrm{V} \
& c)~~ V_2 = 4.73948 ~mathrm{V} \
& c)~~ V_3 = 3.93276 ~mathrm{V} \
& d)~~ V_1 + V_2 + V_3 = V = 12 ~mathrm{V}
end{align*}
$$
$$
I = dfrac{V}{R}
$$
where $V$ is voltage across this resistor. Applying Ohm’s law to resistor $R_2$ would give us:
$$
begin{align*}
I = dfrac{V_2}{R_2} tag{1}
end{align*}
$$
where $V_2$ is voltage across resistor $R_2$ and $I$ is current flowing through it. This current is also unknown, but since $R_1$ and $R_2$ are connected in series, this is the same current $I$ flowing through the battery, which can be calculated by applying the Ohm’s law to the equivalent circuit (with equivalent resistance $R_e$ instead of $R_1$ and $R_2$ separately), stated as:
$$
I = dfrac{V}{R_e}
$$
where $V$ is voltage across the battery and $R_e$ is equivalent resistance of the two resistors connected in series, which we know is calculated as:
$$
begin{align*}
R_e &= R_1 + R_2 \
tag{plug in the values} \
R_e &= 475 ~mathrm{k Omega} + 235 ~mathrm{k Omega} \
R_e &= 710 ~mathrm{k Omega }
end{align*}
$$
$$
begin{align*}
I &= dfrac{V}{R_e} \
tag{plug in the values} \
I &= dfrac{ 45 ~mathrm{V} }{710 ~mathrm{k Omega }} \
tag{$ 1 ~mathrm{k Omega } = 1000 ~mathrm{Omega } $} \
I &= dfrac{ 45 ~mathrm{V} }{710 cdot 1000 ~mathrm{Omega } } \
I &= 6.338 cdot 10^{-5} ~mathrm{A}
end{align*}
$$
Now that we have current $I$ flowing through the voltage divider, we can calculate $V_2$ from equation $(1)$:
$$
begin{align*}
I &= dfrac{V_2}{R_2} \
tag{express $V_2 $ from the equation above } \
V_2 &= I R_2 \
tag{plug in the values} \
V_2 &= 6.338 cdot 10^{-5} ~mathrm{A} cdot 235 ~mathrm{k Omega} \
tag{$ 1 ~mathrm{k Omega } = 1000 ~mathrm{Omega } $} \
V_2 &= 6.338 cdot 10^{-5} ~mathrm{A} cdot 235 cdot 1000 ~mathrm{Omega }
end{align*}
$$
$$
boxed{ V_2 = 14.8944 ~mathrm{V} }
$$
V_2 = 14.8944 ~mathrm{V}
$$
We want the output voltage $V_1$ on resistor $R_1$ to be $V_1 = 2.2 ~mathrm{V}$.
Let’s apply Ohm’s law to resistor $R_1$. Ohm’s law states that current $I$ flowing through a resistor with resistance $R$ is equal to:
$$
I = dfrac{V}{R}
$$
where $V$ is voltage across this resistor. Applying Ohm’s law to resistor $R_1$ would give us:
$$
begin{align*}
I &= dfrac{V_1}{R_1} \
tag{plug in the values} \
I &= dfrac{ 2.2 ~mathrm{V} }{ 1.2 ~mathrm{k Omega} } \
tag{$ 1 ~mathrm{k Omega } = 1000 ~mathrm{Omega } $} \
I &= dfrac{ 2.2 ~mathrm{V} }{ 1.2 cdot 1000 ~mathrm{ Omega} } \
I &= 1.8333 cdot 10^{-3} ~mathrm{A}
end{align*}
$$
$$
I = 1.8333 ~mathrm{mA}
$$
where $V_1$ is voltage across resistor $R_1$ and $I$ is current flowing through it.
$$
begin{align*}
I = dfrac{V}{R_e} tag{1}
end{align*}
$$
where $V$ is voltage across the battery and $R_e$ is equivalent resistance of the two resistors connected in series, which we know is calculated as:
$$
begin{align*}
R_e &= R_1 + R_2
end{align*}
$$
We can plug in this expression for $R_e$ into equation $(1)$
$$
begin{align*}
I &= dfrac{V}{R_e} \
I &= dfrac{V}{R_1 + R_2} \
tag{express $R_1 + R_2$ from the equation above }\
R_1 + R_2 &= dfrac{V}{I} \
tag{express $R_2$ from the equation above } \
R_2 &= dfrac{V}{I} – R_1 \
tag{plug in the values} \
R_2 &= dfrac{ 12 ~mathrm{V} }{ 1.8333 ~mathrm{mA} } – 1.2 ~mathrm{k Omega} \
tag{$ 1 ~mathrm{k Omega } = 1000 ~mathrm{Omega } $} \
tag{$ 1 ~mathrm{mA} = 10 ^{-3} ~mathrm{A} $} \
R_2 &= dfrac{12 ~mathrm{V}}{ 1.8333 cdot 10 ^{-3} ~mathrm{A} } – 1.2 cdot 1000 ~mathrm{Omega }
end{align*}
$$
$$
boxed{ R_2 = 5345.4545 ~mathrm{Omega } }
$$
R_2 = 5345.4545 ~mathrm{Omega }
$$
$a)~~$ To find the equivalent resistance of these resistors connected in parallel, let’s remember that equivalent resistance $R_e$ of three resistors connected in parallel is calculated from:
$$
dfrac{1 }{R_e} = dfrac{1}{R_1} + dfrac{1}{R_2} + dfrac{1}{R_3}
$$
where $R_1$, $R_2$ and $R_3$ are resistances of individual resistors connected in parallel. In our case, all three resistors have the same resistance
$R_1 = R_2 = R_3 = R$. We thus have:
$$
begin{align*}
dfrac{1}{R_e} &= dfrac{1}{R} + dfrac{1}{R} + dfrac{1}{R} \
dfrac{1}{R_e} &= dfrac{3}{R} \
tag{find reciprocal value of the equation above} \
R_e &= dfrac{R}{3} \
tag{plug in the values} \
R_e &= dfrac{15 ~Omega }{3}
end{align*}
$$
$$
boxed{ R_e = 5 ~Omega }
$$
$$
begin{align*}
I &= dfrac{V}{R_e} \
tag{plug in the values} \
I &= dfrac{ 30 ~mathrm{V} }{ 5 ~Omega }
end{align*}
$$
$$
boxed{ I = 6 ~mathrm{A} }
$$
$$
I = dfrac{V}{ R}
$$
We see that we can use the equation above to find the current flowing through each of the resistors.
Current flowing through resistor $R_1 = R$ is equal to:
$$
begin{align*}
I_1 &= dfrac{V}{R} = dfrac{ 30 ~mathrm{V} }{15 ~Omega } = 2 ~mathrm{A}
end{align*}
$$
Current flowing through resistor $R_2 = R$ is equal to:
$$
begin{align*}
I_2 &= dfrac{V}{R} = dfrac{ 30 ~mathrm{V} }{15 ~Omega } = 2 ~mathrm{A}
end{align*}
$$
Current flowing through resistor $R_1 = R$ is equal to:
$$
begin{align*}
I_3 &= dfrac{V}{R} = dfrac{ 30 ~mathrm{V} }{15 ~Omega } = 2 ~mathrm{A}
end{align*}
$$
As expected, since voltage $V$ across each of the resistors is the same and their resistance is the same, we find that the same current flows through each of the resistors:
$$
boxed{ I_1 = I_2 = I_3 = 2 ~mathrm{A} }
$$
begin{align*}
& a)~~ R_e = 5 ~Omega \
& b)~~ I = 6 ~mathrm{A} \
& c)~~ I_1 = I_2 = I_3 = 2 ~mathrm{A} \
end{align*}
$$
$R_1 = 120 ~Omega$,
$R_2 = 60 ~Omega$ and
$R_3 = 40 ~Omega$
connected in parallel across a battery with voltage $V = 12 ~mathrm{V}$.
$a)~~$ To find the equivalent resistance of these resistors connected in parallel, let’s remember that equivalent resistance $R_e$ of three resistors with resistances $R_1$, $R_2$ and $R_3$ connected in parallel is calculated from:
$$
begin{align*}
dfrac{1}{R_e} &= dfrac{1}{R_1} + dfrac{1}{R_2} + dfrac{1}{R_3} \
tag{plug in the values} \
dfrac{1}{R_e} &= dfrac{1}{ 120 ~Omega } + dfrac{1}{60 ~Omega } + dfrac{1}{40 ~Omega } \
tag{common denominator is $120 ~Omega $} \
dfrac{1}{R_e} &= dfrac{1}{ 120 ~Omega } + dfrac{2}{ 2 cdot 60 ~Omega } + dfrac{3}{ 3 cdot 40 ~Omega } \
dfrac{1}{R_e} &= dfrac{1}{120~Omega } + dfrac{2}{ 120 ~Omega } +
dfrac{3}{ 120 ~Omega } \
dfrac{1}{R_e} &= dfrac{6}{120 ~Omega } \
tag{find reciprocal value of the equation above} \
R_e &= dfrac{120 ~Omega }{6}
end{align*}
$$
$$
boxed{ R_e = 20 ~Omega }
$$
$$
begin{align*}
I &= dfrac{V}{R_e} \
tag{plug in the values} \
I &= dfrac{ 12 ~mathrm{V} }{ 20 ~Omega }
end{align*}
$$
$$
boxed{ I = 0.6 ~mathrm{A} }
$$
$$
I = dfrac{V}{ R}
$$
We see that we can use the equation above to find the current flowing through each of the resistors.
Current flowing through resistor $R_1 = 120 ~Omega$ is equal to:
$$
begin{align*}
I_1 &= dfrac{V}{R_1} \
I_1 &= dfrac{ 12 ~mathrm{V} }{120 ~Omega }
end{align*}
$$
$$
boxed{ I_1 = 0.1 ~mathrm{A} }
$$
Current flowing through resistor $R_2 = 60 ~Omega$ is equal to:
$$
begin{align*}
I_2 &= dfrac{V}{R_2} \
I_2 &= dfrac{ 12 ~mathrm{V} }{ 60 ~Omega }
end{align*}
$$
$$
boxed{ I_2 = 0.2 ~mathrm{A} }
$$
Current flowing through resistor $R_1 = 40 ~Omega$ is equal to:
$$
begin{align*}
I_3 &= dfrac{V}{R_3} \
I_3 &= dfrac{ 12 ~mathrm{V} }{ 40 ~Omega }
end{align*}
$$
$$
boxed{ I_3 = 0.3 ~mathrm{A} }
$$
As expected, since resistors $R_1$, $R_2$ and $R_3$ are connected in parallel, total current $I$ in the circuit is equal to a sum of current through individual resistors:
$$
I = I_1 + I_2 + I_3
$$
begin{align*}
& a)~~ R_e = 20 ~Omega \
& b)~~ I = 0.6 ~mathrm{A} \
& c)~~ I_1 = 0.1 ~mathrm{A} \
& c)~~ I_2 = 0.2 ~mathrm{A} \
& c)~~ I_3 = 0.3 ~mathrm{A}
end{align*}
$$
$R_1 = 15 ~Omega$,
$R_2 = 15 ~Omega$ and
$R_3 = 10 ~Omega$ connected in parallel.
$$
begin{align*}
dfrac{1}{R_e} &= dfrac{1}{R_1} + dfrac{1}{R_2} + dfrac{1}{R_3} \
tag{plug in the values} \
dfrac{1}{R_e} &= dfrac{1}{ 15 ~Omega } + dfrac{1}{15 ~Omega } + dfrac{1}{10 ~Omega } \
tag{common denominator is $ 30 ~Omega $} \
dfrac{1}{R_e} &= dfrac{2}{ 30 ~Omega } + dfrac{2}{30 ~Omega } + dfrac{3}{30 ~Omega } \
dfrac{1}{R_e} &= dfrac{7}{30 ~Omega } \
tag{find the reciprocal value of the equation above} \
R_e &= dfrac{30 ~Omega}{7 }
end{align*}
$$
$$
boxed{ R_e = 4.2857 ~Omega }
$$
Going back to Problem 15, we see that if all three resistors were the same and equal $R = 15 ~Omega$, the equivalent resistance is $R_e = 5 ~Omega$. As we can see, when resistor $R_3 = 10 ~Omega$ is used instead of one of the initial resistors, equivalent resistance is $R_e = 4.2857 ~Omega$, which is lower than the initial value from Problem 15.
$$
begin{align*}
I &= dfrac{V}{R_e} \
tag{plug in the values} \
I &= dfrac{ 30 ~mathrm{V} }{ 4.2857 ~Omega }
end{align*}
$$
$$
boxed{ I = 7 ~mathrm{A} }
$$
Going back to problem 15, we see that if all three resistors were the same and equal $R = 15 ~Omega$, current $I$ through the circuit is $I = 6 ~mathrm{A}$. As we can see, when resistor $R_3 = 10 ~Omega$ is used instead of one of the initial resistors, current $I$ through the circuit is $I = 7 ~mathrm{A}$, which is higher than the initial value from Problem 15.
$$
I = dfrac{V}{ R}
$$
We see that we can use the equation above to find the current flowing through each of the resistors.
Current flowing through resistor $R_1 = 15 ~Omega$ is equal to:
$$
begin{align*}
I_1 &= dfrac{V}{R} = dfrac{ 30 ~mathrm{V} }{15 ~Omega } = 2 ~mathrm{A}
end{align*}
$$
Current flowing through resistor $R_2 = 15 ~Omega$ is equal to:
$$
begin{align*}
I_2 &= dfrac{V}{R} = dfrac{ 30 ~mathrm{V} }{15 ~Omega } = 2 ~mathrm{A}
end{align*}
$$
Current flowing through resistor $R_1 = 10 ~Omega$ is equal to:
$$
begin{align*}
I_3 &= dfrac{V}{R} = dfrac{ 30 ~mathrm{V} }{15 ~Omega } = 3 ~mathrm{A}
end{align*}
$$
As expected, current flowing through the $15 ~Omega$ resistors didn’t change. Note that no change was expected, due to a fact that current flowing through a resistor only depends on voltage across it and its resistance, which didn’t change. We see, however, that $I_3$ changed and it changed because current $I_3$ now flows through a $10 ~Omega$ resistor and not the $15 ~Omega$ resistor.
begin{align*}
& a)~~ text{Yes, equivalent resistance decreases to}~ R_e = 4.2857 ~Omega \
& b)~~ text{Yes, current increases to}~ I = 7 ~mathrm{A} \
& c)~~ text{No, current through a $15 ~Omega $ resistor stays the same, }~I_1 = I_2 = 2 ~mathrm{A} \
end{align*}
$$
We essentially need to find the resistance of resistor $R_2$ added to the branch so that resistance of the branch is reduced to the desired value and decide the type of connection needed to reduce the resistance.
Let’s first see what happens when we connect the two resistors with the same resistance $R$ in series. The equivalent resistance $R_{es}$ of their connection in series (resistance of the branch containing both resistors connected in series) is calculated as:
$$
R_{es} = R + R = 2R
$$
We see that connecting the two resistors in series results in increase of overall resistance.
Now let’s see what happens when we connect the same two resistors with resistance $R$ in parallel. Their equivalent resistance $R_{ep}$ of their connection in parallel (resistance of the branch containing both resistors connected in parallel) is calculated from:
$$
begin{align*}
dfrac{1}{R_{ep}} = dfrac{1}{R} + dfrac{1}{R} = dfrac{2}{R} ~~~rightarrow~~~ R_{ep} = dfrac{R}{2}
end{align*}
$$
$$
begin{align*}
dfrac{1}{R_e} &= dfrac{1}{R_1} + dfrac{1}{R_2} \
tag{We can express $R_2$ from the equation above} \
tag{express $dfrac{1}{R_2}$ from the equation above} \
dfrac{1}{R_2} &= dfrac{1}{R_e} – dfrac{1}{R_1} \
tag{common denominator is $R_1 cdot R_e$} \
dfrac{1}{R_2} &= dfrac{R_1}{R_1 R_e} – dfrac{R_e}{R_e R_1} \
dfrac{1}{R_2} &= dfrac{R_1- R_e }{R_1 R_e} \
tag{find the reciprocal value of the equation} \
R_2 &= dfrac{R_1 R_e}{R_1 – R_e} \
tag{plug in the values} \
R_2 &= dfrac{ 150 ~mathrm{Omega } cdot 93 ~mathrm{Omega } }{150 ~mathrm{Omega } – 93 ~mathrm{Omega } }
end{align*}
$$
$$
boxed{ R_2 = 244.7368 ~Omega }
$$
R_2 = 244.7368 ~Omega
$$
We are given:
$$
begin{align*}
text{resistance of the first resistor}~ R_1 = 12 ~mathrm{Omega } \
text{power rating of the first resistor}~ P_1 = 2 ~mathrm{W} \
text{resistance of the second resistor}~ R_1 = 6 ~mathrm{Omega } \
text{power rating of the second resistor}~ P_1 = 4 ~mathrm{W} \
end{align*}
$$
These power ratings correspond to voltage $V$ being applied and currents $I_1$ and $I_2$ flowing through the resistors. We can calculate power rating $P$ on the resistor as:
$$
begin{align*}
P &= I^2 R \
tag{express $I$ from the equation above} \
I^2 &= dfrac{P}{R} \
I &= sqrt{dfrac{P}{R}}
end{align*}
$$
We can use the equation above to calculate the currents $I_1$ and $I_2$ flowing through each resistor.
Current $I_1$ flowing through resistor $R_1$ is equal to:
$$
begin{align*}
I_1 &= sqrt{dfrac{P_1}{R_1}} \
tag{plug in the values} \
I_1 &= sqrt{dfrac{ 2 ~mathrm{W} }{12 ~mathrm{Omega } }} \
I_1 &= 0.4082 ~mathrm{A}
end{align*}
$$
Current $I_2$ flowing through resistor $R_2$ is equal to:
$$
begin{align*}
I_2 &= sqrt{dfrac{P_2}{R_2}} \
tag{plug in the values} \
I_2 &= sqrt{dfrac{ 4 ~mathrm{W} }{6 ~mathrm{Omega } }} \
I_2 &= 0.8165 ~mathrm{A}
end{align*}
$$
$$
I_1 = dfrac{V}{R_1} ~~~text{and}~~~ I_2 = dfrac{V}{R_2}
$$
We see that current is proportional to the voltage across the resistor, which means that if we increase the voltage, current flowing through the resistors will also increase.
Keep in mind that power output, which represents the power output to heat will be equal to:
$$
P_1 = I_1 V ~~~text{and}~~~ P_2 = I_2 V
$$
where $P_1$ is power output on resistor $R_1$ and $P_2$ is power output on resistor $R_2$.
In a given time interval $t$, energy $E$ given off as heat $Q$ on resistor with power rating $P$ is given as:
$$
Q = E = P t
$$
Of course, the more heat the resistor gives off, the hotter it is. This means that temperature $T$ of the resistor is proportional to the heat given off by resistor:
$$
T propto Q
$$
As we can see from the equation above, energy given of as heat $Q$ is proportional to the power rating $P$ of the resistor, which means:
$$
T propto Q ~rightarrow~ T propto P
$$
As said, power rating $P$ can be expressed as a product of voltage $V$ across the resistor and current $I$ flowing through it, which means:
$$
T propto V I
$$
Since voltage across both of the resistors is the same, we are only interested in how current $I$ through a resistor affects the temperature $T$ of the resistor:
$$
T propto I
$$
As we can see, the higher the current through a resistor, the higher the temperature of the resistor. Since we proved that current $I_2$ is always higher than current $I_1$, we conclude that temperature $T_2$ on resistor $R_2$ will be higher than the temperature $T_1$ on resistor $R_1$, stated as:
$$
boxed{ T_2 > T_1 }
$$
